F connects to k with a 3. H connects to k with a 3.

H connects to n with a 5. And then j, k, and n all have a 08 edges

to the sync on the right. So, the primary inputs here are really a,

b and c. The primary outputs here are really, j,

k, and n. and there's little three gray boxes by

each one of these, it's where we're going to fill in all the values.

So, let's just remember that the AT is the longest path from the SRC to the

node. The RAT is 12, the cycle time minus the

longest path from the node to the SNK. And the Slack is the RAT minus the AT.

And for now the problem is small enough, we're just going to look at the mixes,

the minimum and maximum delays, by eye, we're just going to figure it out and

write it down. So, here's the same graph a, b, c, d, e,

f, g, h, j, k, n. SRC on the left, SNK on the right.

we're going to compute AT's, RAT's, and Slacks.

But on this slide, we're going to compute the AT's.

And so, that's going to be the left box on the little gray box by all of these

nodes. And we're just going to compute this.

And let's remember this. This is just the longest path the longest

delay, the longest set of adding them together numbers on the red edges from

the SRC. So, we can do that by eye.

So you know, what is the longest path from the SRC to itself?

And the answer is 0. What's the longest path to the a, b, and

c nodes that have 0 weight edges from the SRC?

Also 0. what are the longest paths to d and e?

Well, there's only one edge to each of them.

So it's, you know, 1 and 2, the numbers on the edges.

what's the longest path to f? Well, now we actually have to calculate

something, you know, because you know, you could go you know, through the d

node, 1 plus 5. you know, or you can go from the b node

with a 4 or you can go from the c node with a one and so, you know, you're

going to get 5 plus 1 is a 6, for node f. what about nodes g and h?

Well, node g, there's only one way to get there.

So, you go from d with a 3. 1 plus 3 is 4.

Node h is actually, it's, it's going to be either 6 plus 4 coming from f, or 2

plus 3 coming from e, so it's going to be the 10.

and similarly we can calculate, you know, the longest way to get to node j is a 7.

Okay? The longest way to get to node k is a 12.

and for that one, you actually have to go from f, right, with the 6 down to h with

the 4 and up with the 2. Right.

That's the worst thing that can happen there.

And the worst way to get to node n is 10 plus 5 is 15.

And then, since all those edges have zero weight to the SNK, the SNK is just the

biggest of those, and so it's 15. And so, those are the AT's.

Longest delays from the SRC. How about the RAT's?

So, the RAT will be the middle number on all of these boxes.

And the RAT's are best thought of as starting from the SNK and going backwards

to the SRC. And the RAT's are just 12 minus the

longest path from the node to the SNK. And so, it's sort of the same thing but

just backwards. You've got to follow the edges in the

other direction. All right.

So what's the RAT for the SNK? It's the cycle time.

What's the RAT for all of the nodes j, k, and n that have a zero weight connection

to the SNK? Also 12, the cycle time.

What about node g? Well, there's only one edge going out of

that. So, you know, the path to the SNK is

length 2. So, 12 minus 2 is 10.

load h, you know there's actually a couple of ways back.

The longest way back to the SNK is 5, 12 minus 5 is 7.

And so, we can just continue this. for node f.